52 cards desk question (girlfriend not good at math)

TripleAce

Hi Guys,
Posting on the math forum as I had a little bit of an argument with my girlfriend tonight with regards to mathematical probability to pick some specific combination of cards in a card deck.


Basically suppose we take a card deck of 52 cards, we shuffle it properly (meaning they are in complete random order), and we pick let’s say 3 random cards to hit a three-of-a-kind (ie 7-7-7) or 5 cards of the same suit (ie. 5 clubs) or 5 cards to hit a straight (7-8-9-10-J).


My version: if you pick the first 3 from the top of the deck (or 5 cards for the flush or straight) you have the exact same probability to hit a three of a kind that if you pick 3 random cards in the middle of the deck.


Her version: If you pick 3 random cards in the middle of the deck you have more chances to hit you three-of-a-kind (or 5 cards for flush of straight) than if you pick the first 3 at the very top of the deck.



EDIT: removing last line of original post as I got scared after I read response #12

hivizman

You could try going through the mathematics, but a lot of people just don't get probability. For example, if you ask some people which is more likely to come out in the Lotto: A = {1 2 3 4 5 6} or B = {5 12 14 21 34 43}, they will say B, because they don't consider that a random choice of numbers is just as likely to come up with A as B - they see the pattern in A and believe that this can't be a random outcome. Your girlfriend is seeing a pattern in the choice of the top three cards from a randomly ordered pack, when this set of three cards is just as randomly distributed as any other set of three cards in the pack.

If you think that the maths will help, then you could try this:

Take the top card from the pack. This can be any of thirteen values (Ace to King). Let's suppose it's an Ace. Then there are three more Aces in the remaining 51 cards. Take the second card from the top of the pack. This can be any of 51 cards, so the probability of drawing an Ace is 3/51 or 1/17. Assuming that it is an Ace, there are now two Aces left out of 50 cards. Take the third card from the top of the pack. The probability of drawing an Ace is 2/50 or 1/25 The overall probability of getting three cards the same is thus 1/17 x 1/25 = 1/425.

Now start with another random pack and choose any card at random. This card can be any of thirteen values (Ace to King). Let's suppose it's an Ace. Then there are three more Aces in the remaining 51 cards. Take a second card at random from the pack. This can be any of 51 cards, so the probability of drawing an Ace is 3/51 or 1/17. Assuming that it is an Ace, there are now two Aces left out of 50 cards. Take a third card at random from the pack. The probability of drawing an Ace is 2/50 or 1/25. The overall probability of getting three cards the same is thus 1/17 x 1/25 = 1/425.

If she still doesn't get it, then ask her which is more likely: that a card drawn at random from a properly randomised pack is the Ace of Spades, or that the top card in a properly randomised pack is the Ace of Spades?

Thomas_S_Hunterson

Women eh:rolleyes:

MathsManiac

You're coming at this all wrong - you gotta think outside the box, here.

Instead of trying to convince her of the truth, you need to shamelessly take advantage of the situation: find out how much more likely she thinks it is, and then start betting.

(And not necessarily for money...)

LeixlipRed

Ah I wouldn't be too hard on her. This is a common enough one. I fecking hate probability and finish my masters in maths in 2 months

hivizman

MathsManiac wrote: »

You're coming at this all wrong - you gotta think outside the box, here.

Instead of trying to convince her of the truth, you need to shamelessly take advantage of the situation: find out how much more likely she thinks it is, and then start betting.

(And not necessarily for money...)

Good idea in principle, but you'd need to play the game far too many times if you want to come out ahead, because the probability of getting three of a kind (let alone a flush or straight) on any single draw is so low.

A simpler version would be to ask your girlfriend which is more likely: that the top two cards in a random pack are the same colour (red or black), or that two cards drawn at random are the same colour? The probability that two cards drawn (without replacing the first card before drawing the second) from a randomly shuffled pack are the same colour is 25/51 (slightly less than 1/2, because, once the first card is drawn, there are only 25 remaining cards of the same colour). If she thinks that the probability of getting two cards the same colour at the top of the pack is significantly less than 1/2, whereas the probability of getting two cards the same colour by drawing at random is 1/2 (or is greater than the other probability), then you are on to a winner, which should pay off quite quickly.

Culchie

It all depends on who's got the biggest tits really :pac:

Pherekydes

Sean_K wrote: »

Women eh:rolleyes:

Culchie wrote: »

It all depends on who's got the biggest tits really :pac:

You two are so funny. I logged on while on hols especially to post this.

Most of my class in college were women. Men don't get probability either: Check out every bookie on race days. :rolleyes:

Thomas_S_Hunterson

Pherekydes wrote: »

You two are so funny. I logged on while on hols especially to post this.

Most of my class in college were women. Men don't get probability either: Check out every bookie on race days. :rolleyes:

Pffft...You'll be claiming that women can drive next:p

Fremen

This reminds me of the time I tried to convince my dad that you're just as likely to get a flush in poker when there are two people playing as you are when there are ten people playing.

His reaction to the maths I showed him was "I don't believe you". :rolleyes:

Pgibson

Fremen wrote: »

This reminds me of the time I tried to convince my dad that you're just as likely to get a flush in poker when there are two people playing as you are when there are ten people playing.

His reaction to the maths I showed him was "I don't believe you". :rolleyes:

His answer may not have been as stupid as you think.

He may have interpreted the statement as: "A person at the table is as likely......".

i.e Any one of the 10 players.

.

Khannie

TripleAce wrote: »

where it is so obvious that I am right

You're right that you're right, but you're also wrong. My proof: You will get no sex tonight because you started this thread.

Fremen

Pgibson wrote: »

His answer may not have been as stupid as you think.

He may have interpreted the statement as: "A person at the table is as likely......".

i.e Any one of the 10 players.

.

Nope, we discussed it at length. He seemed to think "they're taking my outs, so it's less likely that I'll hit them." Irritating. Very, very irritating.